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2x(3x+5)=5x+12
We move all terms to the left:
2x(3x+5)-(5x+12)=0
We multiply parentheses
6x^2+10x-(5x+12)=0
We get rid of parentheses
6x^2+10x-5x-12=0
We add all the numbers together, and all the variables
6x^2+5x-12=0
a = 6; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·6·(-12)
Δ = 313
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{313}}{2*6}=\frac{-5-\sqrt{313}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{313}}{2*6}=\frac{-5+\sqrt{313}}{12} $
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